Mark the wrong statement.

$A$ point particle is acted upon by a restoring force $F = -k x^3$. The time period of oscillation is $T$ when the amplitude is $A$. The time period for an amplitude $2A$ will be

Which of the following functions represents a simple harmonic oscillation?

The displacement of a particle performing $S.H.M.$ is given by $Y = A \cos [\pi(t + \phi)]$. If at $t = 0$,the displacement is $y = 2 \text{ cm}$ and velocity is $v = 2\pi \text{ cm/s}$,the value of amplitude $A$ in $\text{cm}$ is:

$A$ simple harmonic motion is represented by $y = 5(\sin 3\pi t + \sqrt{3} \cos 3\pi t) \ cm$. The amplitude and time period of the motion are:

Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial $(t = 0)$ position of the particle,the radius of the circle,and the angular speed of the rotating particle. For simplicity,the sense of rotation may be fixed to be anticlockwise in every case: ($x$ is in $cm$ and $t$ is in $s$).
$(a)\; x = -2 \sin (3t + \pi/3)$
$(b)\; x = \cos (\pi/6 - t)$
$(c)\; x = 3 \sin (2\pi t + \pi/4)$
$(d)\; x = 2 \cos \pi t$